INTRODUCTION TO MATHEMATICAL METHODS ODL TEST ONE JULY INTAKE 2024

DATE : 14TH SEPTEMBER, 2024

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QUESTION ONE

a. Define
i. singleton set [1]
ii. power set of a given set [1]

Answer:
i. A singleton set is a set containing exactly one element. For example, $\{5\}$ is a singleton set.
ii. The power set of a given set $S$ is the set of all subsets of $S$, including the empty set and $S$ itself. It is denoted by $\mathcal{P}(S)$. For example, if $S = \{1, 2\}$, then $\mathcal{P}(S) = \{ \emptyset, \{1\}, \{2\}, \{1, 2\}$.

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b. Determine whether $*$ is commutative or associative.
i. On $\mathbb{Q}$, define $a * b = 2ab + a$ [4]
ii. On $\mathbb{R}$, define $a * b = (a + b)(ab)$ [4]

Answer:
i.

• Commutativity: Check if $a * b = b * a$. $$a\ast b=2ab+a,b\ast a=2ba+b=2ab+b(\text{since multiplication is commutative in }\mathbb{Q})\mathrm{.}$$

$$a * b = 2ab + a, \quad b * a = 2ba + b = 2ab + b \quad (\text{since multiplication is commutative in $\mathbb{Q}$}).$$ $$a * b - b * a = (2ab + a) - (2ab + b) = a - b.$$ This is not always zero (e.g., $a = 1$, $b = 0$: $1 * 0 = 1$, $0 * 1 = 0$, $1 \neq 0$).
Not commutative.

• Associativity: Check if $(a * b) * c = a * (b * c)$. $$a\ast b=2ab+a,(a\ast b)\ast c=(2ab+a)\ast c=2(2ab+a)c+(2ab+a)$$

$$a * b = 2ab + a, \quad (a * b) * c = (2ab + a) * c = 2(2ab + a)c + (2ab + a) = 4abc + 2ac + 2ab + a.$$ $$$$

$$b\ast c=2bc+b,a\ast (b\ast c)=a\ast (2bc+b)=2a(2bc+b)+a$$

$$b * c = 2bc + b, \quad a * (b * c) = a * (2bc + b) = 2a(2bc + b) + a = 4abc + 2ab + a.$$ These are not equal (e.g., $a = 1$, $b = 1$, $c = 1$: $$

$(1 * 1) * 1 = 3 * 1 = 9$, $\mathrm{}$

$1 * (1 * 1) = 1 * 3 = 7$,

$9 \neq 7$).

Not associative.

ii.

• Commutativity: Check if $a * b = b * a$. $$a\ast b=(a+b)(ab)=a^{2}b+a{b}^2,$$

$$b\ast a=(b+a)(ba)=b^{2}a+b{a}^2$$

$$a * b = (a + b)(ab) = a^2b + ab^2, \quad b * a = (b + a)(ba) = b^2a + ba^2 = a^2b + ab^2 \quad (\text{since multiplication is commutative in $\mathbb{R}$}).$$ Thus, $a * b = b * a$.
Commutative.

• Associativity: Check if $(a * b) * c = a * (b * c)$. $$a\ast b=(a+b)(ab),$$

$$(a\ast b)\ast c=[(a+b)(ab)]\ast c$$

$$a * b = (a + b)(ab), \quad (a * b) * c = [(a + b)(ab)] * c = (ab(a + b)) * c.$$ Let $d = ab(a + b)$, then: $$d * c = (d + c)(dc) = [ab(a + b) + c][ab(a + b)c].$$ $$$$

$$b\ast c=(b+c)(bc),a\ast (b\ast c)=a\ast [(b+c)(bc)]$$

$$b * c = (b + c)(bc), \quad a * (b * c) = a * [(b + c)(bc)] = (a + bc(b + c))(a \cdot bc(b + c)).$$ These are not equal (e.g., $a = 1$, $b = 2$, $c = 3$: $$

$(1 * 2) * 3 = 6 * 3 = 162$,

$1 * (2 * 3) = 1 * 30 = 930$,

$162 \neq 930$).
Not associative.

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c. For the function $f(x) = \frac{3x+2}{x-4}$, determine the domain and range of $f(x)$ and then sketch its graph [7]

Answer:

• Domain: Denominator $x - 4 \neq 0$, so $x \neq 4$. Domain is $x \in \mathbb{R} \setminus \{4\}$ or $(-\infty, 4) \cup (4, \infty)$.

• Range: Let $y = f(x)$, solve for $x$:

  $$y=\frac{3x+2}{x-4}⟹y(x-4)=3x+2$$

$$⟹yx-4y=3x+2$$

$$⟹yx-3x=4y+2$$

$$y = \frac{3x + 2}{x - 4} \implies y(x - 4) = 3x + 2 \implies yx - 4y = 3x + 2 \implies yx - 3x = 4y + 2 \implies x(y - 3) = 4y + 2.$$ $$$$

$$x = \frac{4y + 2}{y - 3}, \quad y \neq 3.$$

Thus, range is $y \in \mathbb{R} \setminus \{3\}$ or $(-\infty, 3) \cup (3, \infty)$.

• Graph Sketch:

  • Vertical asymptote: $x = 4$ (denominator zero).
  • Horizontal asymptote: $y = 3$ (as $x \to \pm \infty$, $f(x) \to 3$).
  • Intercepts:
    • $x$-intercept: Set $f(x) = 0$, so $3x + 2 = 0$, $x = -\frac{2}{3}$. Point: $\left(-\frac{2}{3}, 0\right)$.
    • $y$-intercept: Set $x = 0$, $f(0) = \frac{2}{-4} = -\frac{1}{2}$. Point: $\left(0, -\frac{1}{2}\right)$.
  • Behavior:
    • As $x \to 4^-$, $f(x) \to -\infty$ (e.g., $x = 3.9$, $f(3.9) \approx -13.7$).
    • As $x \to 4^+$, $f(x) \to +\infty$ (e.g., $x = 4.1$, $f(4.1) \approx 14.3$).
    • As $x \to \infty$, $f(x) \to 3^+$ (e.g., $x = 5$, $f(5) = 17 > 3$).
    • As $x \to -\infty$, $f(x) \to 3^-$ (e.g., $x = -5$, $f(-5) = \frac{-13}{-9} \approx 1.44 < 3$).

Graph:

branch ($x > 4$): approaches $y = 3$ from above and $x = 4$ from right going to $+\infty$.

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d. Write the set $B = \{ a_n \mid a_{n+1} = 2a_n, a_1 = 3, n \text{ is a natural number} \}$ in roster form [3]

Answer:
The sequence is defined by $a_1 = 3$, $a_{n+1} = 2a_n$.

• $a_1 = 3$,
• $a_2 = 2 \cdot 3 = 6$,
• $a_3 = 2 \cdot 6 = 12$,
• $a_4 = 2 \cdot 12 = 24$,
• $a_5 = 2 \cdot 24 = 48$,
  and so on.
  Thus, $B = \{3, 6, 12, 24, 48, \dots\}$.
  In roster form: $B = \{3, 6, 12, 24, 48, \ldots\}$.

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QUESTION TWO

a. Give the meaning of
i. Transitive relation [1]
ii. Injective relation [1]

Answer:
i. A transitive relation $R$ on a set $A$ satisfies: if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in A$.
ii. An injective relation $R$ from set $A$ to set $B$ satisfies: for every $b \in B$, there is at most one $a \in A$ such that $(a, b) \in R$. (This means $R$ is "one-to-one" in the sense that no two different elements in $A$ relate to the same element in $B$.)

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b. Let the sets $A$ and $B$ be given as $A = [2, 7]$ and $B = (5, 11)$. Write down the set $A - B$ and show it on the number line [3]

Answer:

• $A - B = \{x \mid x \in A \text{ and } x \notin B\}$.

  • $A = [2, 7]$ includes all $x$ such that $2 \leq x \leq 7$.
  • $B = (5, 11)$ includes all $x$ such that $5 < x < 11$.
    Thus:
  • For $x \in [2, 5]$: $x \in A$ and $x \notin B$ (since $x \leq 5$ not in $B$).
  • For $x \in (5, 7]$: $x \in A$ but $x \in B$ (since $5 < x < 11$), so $x \notin A - B$.
    Therefore, $A - B = [2, 5]$.

• Number line representation:

    0    1    2    3    4    5    6    7    8    9    10   11
    |----|----|----|----|----|----|----|----|----|----|----|
              [========================]   (----------------)SORRY LINE THIS IS STARTING FROM 5 TO 11
            A = [2,7]                         B = (5,11)
            A-B = [2,5] (solid from 2 to 5 inclusive)
  

  • Solid circle at 2 and 5, connected by a solid line.

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c. Show that $\sqrt{3}$ is an irrational number [5]

Answer:
Proof by contradiction. Assume $\sqrt{3}$ is rational. Then $\sqrt{3} = \frac{a}{b}$ where $a, b$ are integers with no common factors (coprime), and $b \neq 0$.

$$\sqrt{3}=\frac{a}{b}$$

$$⟹3=\frac{a^2}{b^2}$$

$$\sqrt{3} = \frac{a}{b} \implies 3 = \frac{a^2}{b^2} \implies a^2 = 3b^2.$$

Thus, $a^2$ is divisible by 3, so $a$ is divisible by 3 (since 3 is prime). Let $a = 3k$ for some integer $k$.

$$(3k{)}^2=3b^2$$

$$⟹9k^2=3b^2$$

$$(3k)^2 = 3b^2 \implies 9k^2 = 3b^2 \implies b^2 = 3k^2.$$

Thus, $b^2$ is divisible by 3, so $b$ is divisible by 3. But then $a$ and $b$ both divisible by 3, contradicting that $\frac{a}{b}$ is in lowest terms. Hence, $\sqrt{3}$ is irrational.

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d. Write the following recurring decimals in the form of $\frac{a}{b}$, where $b \neq 0$
i. $0.23423$ [4] (recurring pattern is "23" starting after "234")
ii. $-2.567$ [4] (recurring pattern is "67")

Answer:
i. Let $x = 0.234232323\ldots$ (repeating "23" after the first three decimals).

• The non-repeating part is "234", so multiply by 1000 to shift it left: $$1000x = 234.232323\ldots$$
• Let $y = 0.232323\ldots$. Then $1000x = 234 + y$.
• Solve for $y$: $$y=0.232323\dots ,100y=23.232323\dots$$

$$⟹100y-y=23$$

$$⟹99y=23$$

$$y = 0.232323\ldots, \quad 100y = 23.232323\ldots \implies 100y - y = 23 \implies 99y = 23 \implies y = \frac{23}{99}.$$

• Substitute: $$1000x=234+\frac{23}{99}$$

$$=\frac{234\times 99+23}{99}$$

$$=\frac{23166+23}{99}$$

$$1000x = 234 + \frac{23}{99} = \frac{234 \times 99 + 23}{99} = \frac{23166 + 23}{99} = \frac{23189}{99}.$$ $$$$

$$x=\frac{23189}{99\times 1000}$$

$$x = \frac{23189}{99 \times 1000} = \frac{23189}{99000}.$$

• Simplify fraction: GCD of 23189 and 99000 is 1 (by Euclidean algorithm), so $\frac{23189}{99000}$ is simplified.

ii. Let $x = -2.5676767\ldots$ (repeating "67" after "5").

• Consider the positive part first: $2.5676767\ldots$.
• Multiply by 10 to shift the non-repeating part ("5"): $$10x_{\text{pos}} = 25.676767\ldots \quad (\text{let } x_{\text{pos}} = 2.5676767\ldots)$$
• Let $z = 0.676767\ldots$. Then $10x_{\text{pos}} = 25 + z$.
• Solve for $z$: $$z=0.676767\dots ,100z=67.676767\dots$$

$$⟹100z-z=67$$

$$⟹99z=67$$

$$z = 0.676767\ldots, \quad 100z = 67.676767\ldots \implies 100z - z = 67 \implies 99z = 67 \implies z = \frac{67}{99}.$$

• Substitute: $$10x_{\text{pos}}=25+\frac{67}{99}$$

$$=\frac{25\times 99+67}{99}$$

$$=\frac{2475+67}{99}$$

$$10x_{\text{pos}} = 25 + \frac{67}{99} = \frac{25 \times 99 + 67}{99} = \frac{2475 + 67}{99} = \frac{2542}{99}.$$ $${}_{}$$

$$x_{\text{pos}}=\frac{2542}{99\times 10}$$

$$x_{\text{pos}} = \frac{2542}{99 \times 10} = \frac{2542}{990}.$$

• Simplify: Divide numerator and denominator by 2: $\frac{1271}{495}$. GCD of 1271 and 495 is 1, so simplified.
• Since original is negative: $x = -\frac{1271}{495}$.

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e. Let $Q \subset R$, simplify $(Q' \cap R')'$ [2]

Answer:

• Assume the universal set is $\mathbb{R}$ (real numbers).
• $Q$ is the set of rational numbers, so $Q' = \mathbb{R} \setminus Q$ (irrational numbers).
• $R = \mathbb{R}$, so $R' = \mathbb{R} \setminus \mathbb{R} = \emptyset$.
• Then $Q' \cap R' = (\mathbb{R} \setminus Q) \cap \emptyset = \emptyset$.
• Thus, $(Q' \cap R')' = \emptyset' = \mathbb{R}$ (complement relative to universal set $\mathbb{R}$).
  Simplified form: $\mathbb{R}$.

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QUESTION THREE

a. Define
i. A function [1]
ii. Complement of the set [1]

Answer:
i. A function $f: A \to B$ is a relation that assigns to each element $a \in A$ exactly one element $b \in B$. It is often denoted by $f(a) = b$.
ii. The complement of a set $A$, denoted $A'$ or $A^c$, is the set of all elements in the universal set that are not in $A$. If universal set is $U$, then $A' = U \setminus A$.

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b. Rationalize the denominator
i. $\frac{3-\sqrt{3}}{2+\sqrt{3}}$ [4]

ii. $\frac{4}{\sqrt{5-3\sqrt{2}}}$ [4]

Answer:
i. Multiply numerator and denominator by conjugate of denominator, $2 - \sqrt{3}$:

$$\frac{3-\sqrt{3}}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{(3-\sqrt{3})(2-\sqrt{3})}{(2{)}^2-(\sqrt{3}{)}^2}$$

$$\frac{3 - \sqrt{3}}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{(3 - \sqrt{3})(2 - \sqrt{3})}{(2)^2 - (\sqrt{3})^2} = \frac{3 \cdot 2 + 3 \cdot (-\sqrt{3}) + (-\sqrt{3}) \cdot 2 + (-\sqrt{3}) \cdot (-\sqrt{3})}{4 - 3}.$$ $$$$

$$\text{Numerator: }6-3\sqrt{3}-2\sqrt{3}+3$$

$$\text{Numerator: } 6 - 3\sqrt{3} - 2\sqrt{3} + 3 = 9 - 5\sqrt{3}, \quad \text{Denominator: } 1.$$

So, $\frac{9 - 5\sqrt{3}}{1} = 9 - 5\sqrt{3}$.

ii. Multiply numerator and denominator by conjugate of denominator, $\sqrt{5 + 3\sqrt{2}}$ (since $(\sqrt{a})^* = \sqrt{a}$ if $a > 0$):

$$\frac{4}{\sqrt{5-3\sqrt{2}}}X\frac{\sqrt{5+3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}}$$

$$=\frac{4\sqrt{5+3\sqrt{2}}}{\sqrt{(5{)}^2-(3\sqrt{2}{)}^2}}$$

$$=\frac{4\sqrt{5+3\sqrt{2}}}{\sqrt{25-18}}$$

$$\frac{4}{\sqrt{5 - 3\sqrt{2}}} \cdot \frac{\sqrt{5 + 3\sqrt{2}}}{\sqrt{5 + 3\sqrt{2}}} = \frac{4 \sqrt{5 + 3\sqrt{2}}}{\sqrt{(5)^2 - (3\sqrt{2})^2}} = \frac{4 \sqrt{5 + 3\sqrt{2}}}{\sqrt{25 - 18}} = \frac{4 \sqrt{5 + 3\sqrt{2}}}{\sqrt{7}}.$$

Rationalize again by multiplying numerator and denominator by $\sqrt{7}$:

$$\frac{4\sqrt{5+3\sqrt{2}}X\sqrt{7}}{\sqrt{7}X\sqrt{7}}$$

$$=\frac{4\sqrt{7(5+3\sqrt{2})}}{7}$$

$$\frac{4 \sqrt{5 + 3\sqrt{2}} \cdot \sqrt{7}}{\sqrt{7} \cdot \sqrt{7}} = \frac{4 \sqrt{7(5 + 3\sqrt{2})}}{7} = \frac{4 \sqrt{35 + 21\sqrt{2}}}{7}.$$

This is the rationalized form.

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c. Sketch the graphs of the following functions
i. $f(x) = |x| - 4$ [3]
ii. $h(x) = \sqrt{3 + 8x}$ [3]

Answer:
i. $f(x) = |x| - 4$:

• Shape: V-shaped, vertex at $(0, -4)$.
• Intercepts:
  • $x$-intercepts: $f(x) = 0 \implies |x| = 4 \implies x = \pm 4$. Points: $(-4, 0)$, $(4, 0)$.
  • $y$-intercept: $f(0) = -4$. Point: $(0, -4)$.
• Behavior: For $x \geq 0$, $f(x) = x - 4$; for $x < 0$, $f(x) = -x - 4$.

Graph:

Note: Vertex at (0, -4), branches with slope 1 for $x > 0$ and slope -1 for $x < 0$.

ii. $h(x) = \sqrt{3 + 8x}$:

• Domain: $3 + 8x \geq 0 \implies x \geq -\frac{3}{8}$.
• Intercepts:
  • $x$-intercept: $h(x) = 0 \implies \sqrt{3 + 8x} = 0 \implies 3 + 8x = 0 \implies x = -\frac{3}{8}$. Point: $\left(-\frac{3}{8}, 0\right)$.
  • $y$-intercept: $x = 0$, $h(0) = \sqrt{3} \approx 1.732$. Point: $(0, \sqrt{3})$.
• Behavior: Increasing for $x \geq -\frac{3}{8}$, starts at $\left(-\frac{3}{8}, 0\right)$ and grows slowly (square root function).

Graph:

Note: Curve starts at $\left(-\frac{3}{8}, 0\right)$ and increases, passing through $(0, \sqrt{3}) \approx (0, 1.732)$.

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d. Determine whether the function below is even or odd or none
$h(x) = x^4 - x^3$ [4]

Answer:

• Check $h(-x)$: $$h(-x) = (-x)^4 - (-x)^3 = x^4 - (-x^3) = x^4 + x^3.$$
• Compare to $h(x)$ and $-h(x)$: $$h(x) = x^4 - x^3, \quad -h(x) = -x^4 + x^3.$$

Since $h(-x) \neq h(x)$ 

(e.g., $x = 1$: $h(-1) = 1 + 1 = 2$, $h(1) = 1 - 1 = 0$) and $h(-x) \neq -h(x)$

(e.g., $x = 1$: $-h(1) = 0$, but $h(-1) = 2 \neq 0$),

the function is neither even nor odd.

OR MAYBE WE DO THIS GUYS

$h\left(x\right)=x^4-x^3$

is even, odd, or neither.

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✏️ Step 1: Recall definitions

• A function is even if: $h(-x)=h\left(x\right)$
• A function is odd if: $h(-x)=-h\left(x\right)$

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🔍 Step 2: Evaluate $h(-x)$

$h(-x)=(-x{)}^4-(-x{)}^3=x^4+x^3$

Compare with original:

• $h\left(x\right)=x^4-x^3$
• $h(-x)=x^4+x^3$

Clearly:

• $h(-x)\ne h\left(x\right)$ → not even
• $h(-x)\ne -h\left(x\right)$ → not odd

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✅ Final Answer:

$\overline{)\text{Neither even nor odd}}$

@Dr. Microbiota

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End of Solutions